# Longest Palindromic Subsequence | DP-12

• Difficulty Level : Medium
• Last Updated : 13 Oct, 2022

Given a sequence, find the length of the longest palindromic subsequence in it.

As another example, if the given sequence is “BBABCBCAB”, then the output should be 7 as “BABCBAB” is the longest palindromic subsequence in it. “BBBBB” and “BBCBB” are also palindromic subsequences of the given sequence, but not the longest ones.
The naive solution for this problem is to generate all subsequences of the given sequence and find the longest palindromic subsequence. This solution is exponential in terms of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently be solved using Dynamic Programming.
1) Optimal Substructure:
Let X[0..n-1] be the input sequence of length n and L(0, n-1) be the length of the longest palindromic subsequence of X[0..n-1].
If last and first characters of X are same, then L(0, n-1) = L(1, n-2) + 2.
Else L(0, n-1) = MAX (L(1, n-1), L(0, n-2)).

Following is a general recursive solution with all cases handled.

```// Every single character is a palindrome of length 1
L(i, i) = 1 for all indexes i in given sequence

// IF first and last characters are not same
If (X[i] != X[j])  L(i, j) =  max{L(i + 1, j),L(i, j - 1)}

// If there are only 2 characters and both are same
Else if (j == i + 1) L(i, j) = 2

// If there are more than two characters, and first and last
// characters are same
Else L(i, j) =  L(i + 1, j - 1) + 2 ```

2) Overlapping Subproblems: Following is a simple recursive implementation of the LPS problem. The implementation simply follows the recursive structure mentioned above.

## C++

 `// C++ program of above approach``#include``using` `namespace` `std;` `// A utility function to get max of two integers``int` `max (``int` `x, ``int` `y) { ``return` `(x > y)? x : y; }` `// Returns the length of the longest palindromic subsequence in seq``int` `lps(``char` `*seq, ``int` `i, ``int` `j)``{``// Base Case 1: If there is only 1 character``if` `(i == j)``    ``return` `1;` `// Base Case 2: If there are only 2``// characters and both are same``if` `(seq[i] == seq[j] && i + 1 == j)``    ``return` `2;` `// If the first and last characters match``if` `(seq[i] == seq[j])``    ``return` `lps (seq, i+1, j-1) + 2;` `// If the first and last characters do not match``return` `max( lps(seq, i, j-1), lps(seq, i+1, j) );``}` `/* Driver program to test above functions */``int` `main()``{``    ``char` `seq[] = ``"GEEKSFORGEEKS"``;``    ``int` `n = ``strlen``(seq);``    ``cout << ``"The length of the LPS is "``         ``<< lps(seq, 0, n-1);``    ``return` `0;``}` `// This code is contributed``// by Akanksha Rai`

## C

 `// C program of above approach``#include``#include` `// A utility function to get max of two integers``int` `max (``int` `x, ``int` `y) { ``return` `(x > y)? x : y; }` `// Returns the length of the longest palindromic subsequence in seq``int` `lps(``char` `*seq, ``int` `i, ``int` `j)``{``   ``// Base Case 1: If there is only 1 character``   ``if` `(i == j)``     ``return` `1;` `   ``// Base Case 2: If there are only 2 characters and both are same``   ``if` `(seq[i] == seq[j] && i + 1 == j)``     ``return` `2;` `   ``// If the first and last characters match``   ``if` `(seq[i] == seq[j])``      ``return` `lps (seq, i+1, j-1) + 2;` `   ``// If the first and last characters do not match``   ``return` `max( lps(seq, i, j-1), lps(seq, i+1, j) );``}` `/* Driver program to test above functions */``int` `main()``{``    ``char` `seq[] = ``"GEEKSFORGEEKS"``;``    ``int` `n = ``strlen``(seq);``    ``printf` `(``"The length of the LPS is %d"``, lps(seq, 0, n-1));``    ``getchar``();``    ``return` `0;``}`

## Java

 `//Java program of above approach` `class` `GFG {` `    ``// A utility function to get max of two integers``    ``static` `int` `max(``int` `x, ``int` `y) {``        ``return` `(x > y) ? x : y;``    ``}``    ``// Returns the length of the longest palindromic subsequence in seq` `    ``static` `int` `lps(``char` `seq[], ``int` `i, ``int` `j) {``    ``// Base Case 1: If there is only 1 character``        ``if` `(i == j) {``            ``return` `1``;``        ``}` `    ``// Base Case 2: If there are only 2 characters and both are same``        ``if` `(seq[i] == seq[j] && i + ``1` `== j) {``            ``return` `2``;``        ``}` `    ``// If the first and last characters match``        ``if` `(seq[i] == seq[j]) {``            ``return` `lps(seq, i + ``1``, j - ``1``) + ``2``;``        ``}` `    ``// If the first and last characters do not match``        ``return` `max(lps(seq, i, j - ``1``), lps(seq, i + ``1``, j));``    ``}`  `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String[] args) {``        ``String seq = ``"GEEKSFORGEEKS"``;``        ``int` `n = seq.length();``        ``System.out.printf(``"The length of the LPS is %d"``, lps(seq.toCharArray(), ``0``, n - ``1``));` `    ``}``}`

## Python3

 `# Python 3 program of above approach` `# A utility function to get max``# of two integers``def` `max``(x, y):``    ``if``(x > y):``        ``return` `x``    ``return` `y``    ` `# Returns the length of the longest``# palindromic subsequence in seq``def` `lps(seq, i, j):``    ` `    ``# Base Case 1: If there is``    ``# only 1 character``    ``if` `(i ``=``=` `j):``        ``return` `1` `    ``# Base Case 2: If there are only 2``    ``# characters and both are same``    ``if` `(seq[i] ``=``=` `seq[j] ``and` `i ``+` `1` `=``=` `j):``        ``return` `2``    ` `    ``# If the first and last characters match``    ``if` `(seq[i] ``=``=` `seq[j]):``        ``return` `lps(seq, i ``+` `1``, j ``-` `1``) ``+` `2` `    ``# If the first and last characters``    ``# do not match``    ``return` `max``(lps(seq, i, j ``-` `1``),``               ``lps(seq, i ``+` `1``, j))` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``seq ``=` `"GEEKSFORGEEKS"``    ``n ``=` `len``(seq)``    ``print``(``"The length of the LPS is"``,``                  ``lps(seq, ``0``, n ``-` `1``))``    ` `# This code contributed by Rajput-Ji`

## C#

 `// C# program of the above approach``using` `System;``                    ` `public` `class` `GFG{` `    ``// A utility function to get max of two integers``    ``static` `int` `max(``int` `x, ``int` `y) {``        ``return` `(x > y) ? x : y;``    ``}``    ``// Returns the length of the longest palindromic subsequence in seq`` ` `    ``static` `int` `lps(``char` `[]seq, ``int` `i, ``int` `j) {``    ``// Base Case 1: If there is only 1 character``        ``if` `(i == j) {``            ``return` `1;``        ``}`` ` `    ``// Base Case 2: If there are only 2 characters and both are same``        ``if` `(seq[i] == seq[j] && i + 1 == j) {``            ``return` `2;``        ``}`` ` `    ``// If the first and last characters match``        ``if` `(seq[i] == seq[j]) {``            ``return` `lps(seq, i + 1, j - 1) + 2;``        ``}`` ` `    ``// If the first and last characters do not match``        ``return` `max(lps(seq, i, j - 1), lps(seq, i + 1, j));``    ``}`` ` ` ` `    ``/* Driver program to test above function */``    ``public` `static` `void` `Main() {``        ``String seq = ``"GEEKSFORGEEKS"``;``        ``int` `n = seq.Length;``        ``Console.Write(``"The length of the LPS is "``+lps(seq.ToCharArray(), 0, n - 1));`` ` `    ``}``}` `// This code is contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``

Output

`The length of the LPS is 5`

Considering the above implementation, the following is a partial recursion tree for a sequence of length 6 with all different characters.

```               L(0, 5)
/        \
/          \
L(1,5)          L(0,4)
/    \            /    \
/      \          /      \
L(2,5)    L(1,4)  L(1,4)  L(0,3)```

In the above partial recursion tree, L(1, 4) is being solved twice. If we draw the complete recursion tree, then we can see that there are many subproblems that are solved again and again. Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So LPS problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of the same subproblems can be avoided by constructing a temporary array L[][] in a bottom-up manner.
3) Dynamic Programming Solution:

## C++

 `// A Dynamic Programming based C++ program for LPS problem``// Returns the length of the longest palindromic subsequence in seq``#include``#include` `// A utility function to get max of two integers``int` `max (``int` `x, ``int` `y) { ``return` `(x > y)? x : y; }` `// Returns the length of the longest palindromic subsequence in seq``int` `lps(``char` `*str)``{``   ``int` `n = ``strlen``(str);``   ``int` `i, j, cl;``   ``int` `L[n][n];  ``// Create a table to store results of subproblems`  `   ``// Strings of length 1 are palindrome of length 1``   ``for` `(i = 0; i < n; i++)``      ``L[i][i] = 1;` `    ``// Build the table. Note that the lower diagonal values of table are``    ``// useless and not filled in the process. The values are filled in a``    ``// manner similar to Matrix Chain Multiplication DP solution (See``    ``// https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/). cl is length of``    ``// substring``    ``for` `(cl=2; cl<=n; cl++)``    ``{``        ``for` `(i=0; i

## Java

 `// A Dynamic Programming based Java``// Program for the Egg Dropping Puzzle``class` `LPS``{` `    ``// A utility function to get max of two integers``    ``static` `int` `max (``int` `x, ``int` `y) { ``return` `(x > y)? x : y; }``    ` `    ``// Returns the length of the longest``    ``// palindromic subsequence in seq``    ``static` `int` `lps(String seq)``    ``{``    ``int` `n = seq.length();``    ``int` `i, j, cl;``    ``// Create a table to store results of subproblems``    ``int` `L[][] = ``new` `int``[n][n];``    ` `    ``// Strings of length 1 are palindrome of length 1``    ``for` `(i = ``0``; i < n; i++)``        ``L[i][i] = ``1``;``            ` `        ``// Build the table. Note that the lower``        ``// diagonal values of table are``        ``// useless and not filled in the process.``        ``// The values are filled in a manner similar``        ``//  to Matrix Chain Multiplication DP solution (See``        ``// https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/).``        ``// cl is length of substring``        ``for` `(cl=``2``; cl<=n; cl++)``        ``{``            ``for` `(i=``0``; i

## Python3

 `/``/` `A Dynamic Programming based Java``/``/` `Program ``for` `the Egg Dropping Puzzle``class` `LPS``{` `    ``/``/` `A utility function to get ``max` `of two integers``    ``static ``int` `max` `(``int` `x, ``int` `y) { ``return` `(x > y)? x : y; }``    ` `    ``/``/` `Returns the length of the longest``    ``/``/` `palindromic subsequence ``in` `seq``    ``static ``int` `lps(String seq)``    ``{``    ``int` `n ``=` `seq.length();``    ``int` `i, j, cl;``    ``/``/` `Create a table to store results of subproblems``    ``int` `L[][] ``=` `new ``int``[n][n];``    ` `    ``/``/` `Strings of length ``1` `are palindrome of length ``1``    ``for` `(i ``=` `0``; i < n; i``+``+``)``        ``L[i][i] ``=` `1``;``            ` `        ``/``/` `Build the table. Note that the lower``        ``/``/` `diagonal values of table are``        ``/``/` `useless ``and` `not` `filled ``in` `the process.``        ``/``/` `The values are filled ``in` `a manner similar``        ``/``/`  `to Matrix Chain Multiplication DP solution (See``        ``/``/` `https:``/``/``www.geeksforgeeks.org``/``matrix``-``chain``-``multiplication``-``dp``-``8``/``).``        ``/``/` `cl ``is` `length of substring``        ``for` `(cl``=``2``; cl<``=``n; cl``+``+``)``        ``{``            ``for` `(i``=``0``; i

## C#

 `// A Dynamic Programming based C# Program``// for the Egg Dropping Puzzle``using` `System;` `class` `GFG {` `    ``// A utility function to get max of``    ``// two integers``    ``static` `int` `max (``int` `x, ``int` `y)``    ``{``        ``return` `(x > y)? x : y;``    ``}``    ` `    ``// Returns the length of the longest``    ``// palindromic subsequence in seq``    ``static` `int` `lps(``string` `seq)``    ``{``    ``int` `n = seq.Length;``    ``int` `i, j, cl;``    ` `    ``// Create a table to store results``    ``// of subproblems``    ``int` `[,]L = ``new` `int``[n,n];``    ` `    ``// Strings of length 1 are``    ``// palindrome of length 1``    ``for` `(i = 0; i < n; i++)``        ``L[i,i] = 1;``            ` `        ``// Build the table. Note that the``        ``// lower diagonal values of table``        ``// are useless and not filled in``        ``// the process. The values are``        ``// filled in a manner similar to``        ``// Matrix Chain Multiplication DP``        ``// solution (See``        ``// https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/``        ``// cl is length of substring``        ``for` `(cl = 2; cl <= n; cl++)``        ``{``            ``for` `(i = 0; i < n-cl+1; i++)``            ``{``                ``j = i + cl - 1;``                ` `                ``if` `(seq[i] == seq[j] &&``                                  ``cl == 2)``                    ``L[i,j] = 2;``                ``else` `if` `(seq[i] == seq[j])``                    ``L[i,j] = L[i+1,j-1] + 2;``                ``else``                    ``L[i,j] =``                     ``max(L[i,j-1], L[i+1,j]);``            ``}``        ``}``            ` `        ``return` `L[0,n-1];``    ``}``        ` `    ``/* Driver program to test above``    ``functions */``    ``public` `static` `void` `Main()``    ``{``        ``string` `seq = ``"GEEKSFORGEEKS"``;``        ``int` `n = seq.Length;``        ``Console.Write(``"The length of the "``                  ``+ ``"LPS is "``+ lps(seq));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ` \$y)? \$x : \$y; }` `// Returns the length of the``// longest palindromic``// subsequence in seq``function` `lps(``\$str``)``{``\$n` `= ``strlen``(``\$str``);``\$i``; ``\$j``; ``\$cl``;` `// Create a table to store``// results of subproblems``\$L``[][] = ``array``(``array``());`  `// Strings of length 1 are``// palindrome of length 1``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``\$L``[``\$i``][``\$i``] = 1;` `    ``// Build the table. Note that``    ``// the lower diagonal values``    ``// of table are useless and``    ``// not filled in the process.``    ``// The values are filled in a``    ``// manner similar to Matrix``    ``// Chain Multiplication DP``    ``// solution (See``    ``// https://www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/).``    ``// cl is length of substring``    ``for` `(``\$cl` `= 2; ``\$cl` `<= ``\$n``; ``\$cl``++)``    ``{``        ``for` `(``\$i` `= 0; ``\$i` `< ``\$n` `- ``\$cl` `+ 1; ``\$i``++)``        ``{``            ``\$j` `= ``\$i` `+ ``\$cl` `- 1;``            ``if` `(``\$str``[``\$i``] == ``\$str``[``\$j``] &&``                            ``\$cl` `== 2)``            ``\$L``[``\$i``][``\$j``] = 2;``            ``else` `if` `(``\$str``[``\$i``] == ``\$str``[``\$j``])``            ``\$L``[``\$i``][``\$j``] = ``\$L``[``\$i` `+ 1][``\$j` `- 1] + 2;``            ``else``            ``\$L``[``\$i``][``\$j``] = max(``\$L``[``\$i``][``\$j` `- 1],``                             ``\$L``[``\$i` `+ 1][``\$j``]);``        ``}``    ``}` `    ``return` `\$L``[0][``\$n` `- 1];``}` `// Driver Code``\$seq` `= ``'GEEKSFORGEEKS'``;``\$n` `= ``strlen``(``\$seq``);``echo` `"The length of the "` `.``      ``"LPS is "``, lps(``\$seq``);` `// This code is contributed``// by shiv_bhakt.``?>`

## Javascript

 ``

Output

`The length of the LPS is 5`

Time Complexity: O(n^2), which is much better than the worst-case time complexity of Naive Recursive implementation.

Auxiliary Space: O(n^2),  Creating a table

Using Memoization Technique of Dynamic programming: The idea used here is to reverse the given input string and check the length of the longest common subsequence. That would be the answer for the longest palindromic subsequence.

## C++

 `// A Dynamic Programming based C++ program for LPS problem``// Returns the length of the longest palindromic subsequence``// in seq``#include ``using` `namespace` `std;` `int` `dp[1001][1001];` `// Returns the length of the longest palindromic subsequence``// in seq``int` `lps(string& s1, string& s2, ``int` `n1, ``int` `n2)``{``    ``if` `(n1 == 0 || n2 == 0) {``        ``return` `0;``    ``}``    ``if` `(dp[n1][n2] != -1) {``        ``return` `dp[n1][n2];``    ``}``    ``if` `(s1[n1 - 1] == s2[n2 - 1]) {``        ``return` `dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1);``    ``}``    ``else` `{``        ``return` `dp[n1][n2] = max(lps(s1, s2, n1 - 1, n2),``                                ``lps(s1, s2, n1, n2 - 1));``    ``}``}` `/* Driver program to test above functions */``int` `main()``{``    ``string seq = ``"GEEKSFORGEEKS"``;``    ``int` `n = seq.size();``    ``dp[n][n];``    ``memset``(dp, -1, ``sizeof``(dp));``    ``string s2 = seq;``    ``reverse(s2.begin(), s2.end());``    ``cout << ``"The length of the LPS is "``         ``<< lps(s2, seq, n, n) << endl;``    ``return` `0;``}` `// This code is contributed by Arun Bang`

## Java

 `// Java program of above approach``import` `java.util.*;``class` `GFG {` `    ``// A utility function to get max of two integers``    ``static` `int` `max(``int` `x, ``int` `y) { ``return` `(x > y) ? x : y; }` `    ``// Returns the length of the longest palindromic``    ``// subsequence in seq``    ``static` `int` `lps(``char` `seq[], ``int` `i, ``int` `j, ``int` `dp[][])``    ``{` `        ``// Base Case 1: If there is only 1 character``        ``if` `(i == j) {``            ``return` `dp[i][j] = ``1``;``        ``}` `        ``// Base Case 2: If there are only 2 characters and``        ``// both are same``        ``if` `(seq[i] == seq[j] && i + ``1` `== j) {``            ``return` `dp[i][j] = ``2``;``        ``}``        ``// Avoid extra call for already calculated subproblems, Just return the saved ans from cache``          ``if``(dp[i][j] != -``1``){``          ``return` `dp[i][j];``        ``}``        ``// If the first and last characters match``        ``if` `(seq[i] == seq[j]) {``            ``return` `dp[i][j] = lps(seq, i + ``1``, j - ``1``, dp) + ``2``;``        ``}` `        ``// If the first and last characters do not match``        ``return` `dp[i][j] = max(lps(seq, i, j - ``1``, dp), lps(seq, i + ``1``, j, dp));``    ``}` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String seq = ``"GEEKSFORGEEKS"``;``        ``int` `n = seq.length();``          ``int` `dp[][] = ``new` `int``[n][n];``          ``for``(``int``[] arr: dp)``          ``Arrays.fill(arr, -``1``);``        ``System.out.printf(``"The length of the LPS is %d"``,``                          ``lps(seq.toCharArray(), ``0``, n - ``1``, dp));``    ``}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# A Dynamic Programming based Python program for LPS problem``# Returns the length of the longest palindromic subsequence``# in seq` `dp ``=` `[[``-``1` `for` `i ``in` `range``(``1001``)]``for` `j ``in` `range``(``1001``)]` `# Returns the length of the longest palindromic subsequence``# in seq`  `def` `lps(s1, s2, n1, n2):` `    ``if` `(n1 ``=``=` `0` `or` `n2 ``=``=` `0``):``        ``return` `0` `    ``if` `(dp[n1][n2] !``=` `-``1``):``        ``return` `dp[n1][n2]` `    ``if` `(s1[n1 ``-` `1``] ``=``=` `s2[n2 ``-` `1``]):``        ``dp[n1][n2] ``=` `1` `+` `lps(s1, s2, n1 ``-` `1``, n2 ``-` `1``)``        ``return` `dp[n1][n2]``    ``else``:``        ``dp[n1][n2] ``=` `max``(lps(s1, s2, n1 ``-` `1``, n2), lps(s1, s2, n1, n2 ``-` `1``))``        ``return` `dp[n1][n2]` `# Driver program to test above functions`  `seq ``=` `"GEEKSFORGEEKS"``n ``=` `len``(seq)` `s2 ``=` `seq``s2 ``=` `s2[::``-``1``]``print``(f``"The length of the LPS is {lps(s2, seq, n, n)}"``)` `# This code is contributed by shinjanpatra`

## C#

 `// C# code to implement the approach``using` `System;``using` `System.Numerics;``using` `System.Collections.Generic;` `public` `class` `GFG {` `    ``// A utility function to get max of two integers``    ``static` `int` `max(``int` `x, ``int` `y) { ``return` `(x > y) ? x : y; }` `    ``// Returns the length of the longest palindromic``    ``// subsequence in seq``    ``static` `int` `lps(``char``[] seq, ``int` `i, ``int` `j)``    ``{` `        ``// Base Case 1: If there is only 1 character``        ``if` `(i == j) {``            ``return` `1;``        ``}` `        ``// Base Case 2: If there are only 2 characters and``        ``// both are same``        ``if` `(seq[i] == seq[j] && i + 1 == j) {``            ``return` `2;``        ``}` `        ``// If the first and last characters match``        ``if` `(seq[i] == seq[j]) {``            ``return` `lps(seq, i + 1, j - 1) + 2;``        ``}` `        ``// If the first and last characters do not match``        ``return` `max(lps(seq, i, j - 1), lps(seq, i + 1, j));``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `seq = ``"GEEKSFORGEEKS"``;``        ``int` `n = seq.Length;``        ``Console.Write(``"The length of the LPS is "``                      ``+ lps(seq.ToCharArray(), 0, n - 1));``    ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``

Output

`The length of the LPS is 5`

Time Complexity: O(n2)
Auxiliary Space: O(n2)

Print Longest Palindromic Subsequence
Longest palindrome subsequence with O(n) space